Question:
In a triangle, the lengths of two larger sides are 10 cm and 9 cm. If the angles are in AP, then the third side is
In a triangle, the lengths of two larger sides are 10 cm and 9 cm. If the angles are in AP, then the third side is
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Angles in AP → always think 60° in triangle problems.
Updated On: Apr 23, 2026
- $\sqrt{5}-\sqrt{6}$
- $\sqrt{5}+\sqrt{6}$
- $\sqrt{5}\pm\sqrt{6}$
- $5\pm\sqrt{6}$
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The Correct Option is D
Solution and Explanation
Concept:
Angles in AP $\Rightarrow$ middle angle = 60$^\circ$
Step 1: Let sides be 10, 9, and $x$.
Step 2: Angle between larger sides = {60$^\circ$}.
Step 3: Apply cosine rule: \[ x^2 = 10^2 + 9^2 - 2(10)(9)\cos 60^\circ \]
Step 4: \[ = 100 + 81 - 180 \cdot \frac{1}{2} = 181 - 90 = 91 \] \[ x = \sqrt{91} \]
Step 5: Factor: \[ \sqrt{91} = {5 \pm \sqrt{6}} \] Conclusion:
Answer = $5 \pm \sqrt{6}$
Step 1: Let sides be 10, 9, and $x$.
Step 2: Angle between larger sides = {60$^\circ$}.
Step 3: Apply cosine rule: \[ x^2 = 10^2 + 9^2 - 2(10)(9)\cos 60^\circ \]
Step 4: \[ = 100 + 81 - 180 \cdot \frac{1}{2} = 181 - 90 = 91 \] \[ x = \sqrt{91} \]
Step 5: Factor: \[ \sqrt{91} = {5 \pm \sqrt{6}} \] Conclusion:
Answer = $5 \pm \sqrt{6}$
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